Cho \(A = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{2009}};B = \frac{{2008}}{1} + \frac{{2007}}{2} + \frac{{2006}}{3} + \ldots + \frac{1}{{2008}}\). Tính \(\frac{A}{B}\)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} B = 2018 + \frac{{2007}}{2} + \frac{{2006}}{3} + \ldots + \frac{1}{{2008}}\\ = \left( {1 + \frac{{2007}}{2}} \right) + \left( {1 + \frac{{2006}}{3}} \right) + \ldots + \left( {1 + \frac{1}{{2008}}} \right) + 1 = \frac{{2009}}{2} + \frac{{2009}}{3} + \frac{{2009}}{4} + \ldots + \frac{{2009}}{{2008}} + \frac{{2009}}{{2009}}\\ = 2009\left( {\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{2008}} + \frac{1}{{2009}}} \right) = 2009.A\\ \Rightarrow \frac{A}{B} = \frac{A}{{2009.A}} = \frac{1}{{2009}} \end{array}\)